Problems 1-7 refer to the partially completed table of the six basic solutions of the e -system 2 x 1 + 5 x 2 + s 1 = 32 x 1 + 2 x 2 + s 2 = 14 x 1 x 2 s 1 s 2 A 0 0 32 14 B 0 6.4 0 1.2 C 0 7 − 3 0 D 16 0 0 − 2 E 0 0 F 0 0 In basic solution B , which variables are basic?

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Answer 1

Textbook Question

Chapter 6, Problem 1RE

Problems 1-7 refer to the partially completed table of the six basic solutions of the e -system

2 x 1 + 5 x 2 + s 1 = 32 x 1 + 2 x 2 + s 2 = 14 x 1 x 2 s 1 s 2 A 0 0 32 14 B 0 6.4 0 1.2 C 0 7 − 3 0 D 16 0 0 − 2 E 0 0 F 0 0

In basic solution B , which variables are basic?

Expert Solution & Answer

To determine

The variables which are basic from the basic solution B of the e-system 2x1+5x2+s1 =32x1+2x2 +s2=14 given in the table below.

x1x2s1s2A003214B06.401.2C07−30D1600−2E00F00

Answer to Problem 1RE

The basic variables are x2,s2.

Explanation of Solution

Consider the given e-system.

2x1+5x2+s1 =32x1+2x2 +s2=14

The variables that have been assigned the value zero are non-basic variables and others that have not been assigned the value zero are termed as basic variables.

Basic solution B of the e-system is,

x1x2s1s206.401.2

Thus, from the basic solution of the e-system, x2,s2 have not been assigned the value zero.

Therefore, the basic variables are x2,s2.

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Answer 2

Textbook Question

Chapter 6, Problem 1RE

Problems 1-7 refer to the partially completed table of the six basic solutions of the e -system

2 x 1 + 5 x 2 + s 1 = 32 x 1 + 2 x 2 + s 2 = 14 x 1 x 2 s 1 s 2 A 0 0 32 14 B 0 6.4 0 1.2 C 0 7 − 3 0 D 16 0 0 − 2 E 0 0 F 0 0

In basic solution B , which variables are basic?

Expert Solution & Answer

To determine

The variables which are basic from the basic solution B of the e-system 2x1+5x2+s1 =32x1+2x2 +s2=14 given in the table below.

x1x2s1s2A003214B06.401.2C07−30D1600−2E00F00

Answer to Problem 1RE

The basic variables are x2,s2.

Explanation of Solution

Consider the given e-system.

2x1+5x2+s1 =32x1+2x2 +s2=14

The variables that have been assigned the value zero are non-basic variables and others that have not been assigned the value zero are termed as basic variables.

Basic solution B of the e-system is,

x1x2s1s206.401.2

Thus, from the basic solution of the e-system, x2,s2 have not been assigned the value zero.

Therefore, the basic variables are x2,s2.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 6, Problem 1RE

Problems 1-7 refer to the partially completed table of the six basic solutions of the e -system

2 x 1 + 5 x 2 + s 1 = 32 x 1 + 2 x 2 + s 2 = 14 x 1 x 2 s 1 s 2 A 0 0 32 14 B 0 6.4 0 1.2 C 0 7 − 3 0 D 16 0 0 − 2 E 0 0 F 0 0

In basic solution B , which variables are basic?

Expert Solution & Answer

To determine

The variables which are basic from the basic solution B of the e-system 2x1+5x2+s1 =32x1+2x2 +s2=14 given in the table below.

x1x2s1s2A003214B06.401.2C07−30D1600−2E00F00

Answer to Problem 1RE

The basic variables are x2,s2.

Explanation of Solution

Consider the given e-system.

2x1+5x2+s1 =32x1+2x2 +s2=14

The variables that have been assigned the value zero are non-basic variables and others that have not been assigned the value zero are termed as basic variables.

Basic solution B of the e-system is,

x1x2s1s206.401.2

Thus, from the basic solution of the e-system, x2,s2 have not been assigned the value zero.

Therefore, the basic variables are x2,s2.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 6, Problem 1RE

Problems 1-7 refer to the partially completed table of the six basic solutions of the e -system

2 x 1 + 5 x 2 + s 1 = 32 x 1 + 2 x 2 + s 2 = 14 x 1 x 2 s 1 s 2 A 0 0 32 14 B 0 6.4 0 1.2 C 0 7 − 3 0 D 16 0 0 − 2 E 0 0 F 0 0

In basic solution B , which variables are basic?

Expert Solution & Answer

To determine

The variables which are basic from the basic solution B of the e-system 2x1+5x2+s1 =32x1+2x2 +s2=14 given in the table below.

x1x2s1s2A003214B06.401.2C07−30D1600−2E00F00

Answer to Problem 1RE

The basic variables are x2,s2.

Explanation of Solution

Consider the given e-system.

2x1+5x2+s1 =32x1+2x2 +s2=14

The variables that have been assigned the value zero are non-basic variables and others that have not been assigned the value zero are termed as basic variables.

Basic solution B of the e-system is,

x1x2s1s206.401.2

Thus, from the basic solution of the e-system, x2,s2 have not been assigned the value zero.

Therefore, the basic variables are x2,s2.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The variables which are basic from the basic solution B of the e-system 2x1+5x2+s1 =32x1+2x2 +s2=14 given in the table below.

x1x2s1s2A003214B06.401.2C07−30D1600−2E00F00

Answer to Problem 1RE

The basic variables are x2,s2.

Explanation of Solution

Consider the given e-system.

2x1+5x2+s1 =32x1+2x2 +s2=14

The variables that have been assigned the value zero are non-basic variables and others that have not been assigned the value zero are termed as basic variables.

Basic solution B of the e-system is,

x1x2s1s206.401.2

Thus, from the basic solution of the e-system, x2,s2 have not been assigned the value zero.

Therefore, the basic variables are x2,s2.

Answer 8

Expert Solution & Answer

To determine

The variables which are basic from the basic solution B of the e-system 2x1+5x2+s1 =32x1+2x2 +s2=14 given in the table below.

x1x2s1s2A003214B06.401.2C07−30D1600−2E00F00

Answer to Problem 1RE

The basic variables are x2,s2.

Explanation of Solution

Consider the given e-system.

2x1+5x2+s1 =32x1+2x2 +s2=14

The variables that have been assigned the value zero are non-basic variables and others that have not been assigned the value zero are termed as basic variables.

Basic solution B of the e-system is,

x1x2s1s206.401.2

Thus, from the basic solution of the e-system, x2,s2 have not been assigned the value zero.

Therefore, the basic variables are x2,s2.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The variables which are basic from the basic solution B of the e-system 2x1+5x2+s1 =32x1+2x2 +s2=14 given in the table below.

x1x2s1s2A003214B06.401.2C07−30D1600−2E00F00

Answer 12

Answer to Problem 1RE

The basic variables are x2,s2.

Answer 13

Explanation of Solution

Consider the given e-system.

2x1+5x2+s1 =32x1+2x2 +s2=14

The variables that have been assigned the value zero are non-basic variables and others that have not been assigned the value zero are termed as basic variables.

Basic solution B of the e-system is,

x1x2s1s206.401.2

Thus, from the basic solution of the e-system, x2,s2 have not been assigned the value zero.

Therefore, the basic variables are x2,s2.

Answer 14

Ch. 6.1 - The following linear programming problem has only...Ch. 6.1 - Use the table method to solve the following linear...Ch. 6.1 - Use the table method to solve the following linear...Ch. 6.1 - Refer to Example 1. Find the basic solution for...Ch. 6.1 - Construct the table of basic solutions and use it...Ch. 6.1 - Construct the table of basic solutions and use it...Ch. 6.1 - Refer to Table 5. For the basic solution...Ch. 6.1 - In Problems 1-8, evaluate the expression. (If...Ch. 6.1 - In Problems 1-8, evaluate the expression. (If...Ch. 6.1 - In Problems 1-8, evaluate the expression. (If...

Problems 1-7 refer to the partially completed table of the six basic solutions of the e -system 2 x 1 + 5 x 2 + s 1 = 32 x 1 + 2 x 2 + s 2 = 14 x 1 x 2 s 1 s 2 A 0 0 32 14 B 0 6.4 0 1.2 C 0 7 − 3 0 D 16 0 0 − 2 E 0 0 F 0 0 In basic solution B , which variables are basic?

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Answer to Problem 1RE

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